CHEM1001
2014-J-7
June 2014
• Complete the Lewis structure of formic acid below by adding double bond(s) and
lone pair(s).
Formic acid can form dimers in which two molecules are paired by mutual hydrogen
bonding. Draw a dimer of formic acid, clearly showing the hydrogen bonds between
the molecules.
Formic acid may lose H+ from the oxygen to give a formate ion, shown in Equation 1.
HCOOH(aq)
HCOO–(aq) + H+(aq)
Equation 1
Draw a Lewis structure of the formate ion and use it to illustrate the concept of
resonance.
Comment on the carbon-oxygen bond lengths in formic acid and the carbon-oxygen
bond lengths in the formate ion.
In formic acid, the C=O double bond is shorter than the C–O bond. In the
formate ion, the two C–O bonds are identical because of resonance and hence
are the same length.
THIS QUESTION IS CONTINUED ON THE NEXT PAGE.
22/01(a)
Marks
5
CHEM1001
2013-J-7
June 2013
22/01(a)
Marks
• By adding double bonds and lone pairs, complete the structural formulae of the
nitrogen bases adenine and thymine below.
H
5
H
N
C
N
N
C
C
C
H
C
H
H
H
N
N
H
H
C
C
N
H
C
H
C
H
H
C
O
N
O
N
H
C
N
N
C
C
C
H
C
thymine
H
N
N
H
adenine
In DNA, these two molecules interact through two hydrogen bonds. Redraw the
structures below showing the alignment of the two molecules that allows this to occur
and clearly show the hydrogen bonds.
H
H
H
N
N
H
H
C
H
C
C
N
N
C
C
C
N
O
H
C
N
H
C
C
C
N
H
O
H
H
CHEM1001
2013-J-11
June 2013
• The boiling point of NH3 is –33 °C and that of HF is +20 °C. Explain this difference
in terms of the strengths of the intermolecular forces between these molecules.
The strongest intermolecular force in both comes from hydrogen bonding.
Each HF molecule possesses 3 lone pairs on F and 1 H. HF molecules on average
make 2 H-bonds.
Each NH3 molecule possesses 1 lone pair on N and 3 H. NH3 molecules on
average also make 2 H-bonds.
As fluorine is more electronegative than nitrogen, the H-F bonds are much more
polar than the N-H bonds. Due to the higher partial charges on H and F in HF, a
hydrogen bond between HF molecules is stronger than that between NH3
molecules.
The higher boiling point of HF is thus due to stronger H-bonds.
Explain why the boiling point of water (100 °C) is higher than both HF and NH3.
The polarity of the O-H bonds in H2O is intermediate between that of H-F and NH bonds. It is expected that the individual H-bonds between H2O molecules will
also be intermediate in strength.
Each H2O molecule possesses 2 lone pairs on O and 2 H. H2O molecules are thus
able to form an average of 4 H-bonds.
H2O has a higher boiling point than NH3 because (i) the H-bonds are stronger
and (ii) it contains twice as many H-bonds.
H2O has a higher boiling point than HF because it contains twice as many Hbonds, despite these being individually weaker.
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.
22/01(a)
Marks
3
CHEM1001
2012-J-11
June 2012
• Rank the following compounds in order of increasing boiling point? Justify your
answer.
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Marks
3
CH3CH2OCH2CH3, CH3OH, CH4, CH3CH3, CH3CH2OH
CH4 < CH3CH3 < CH3CH2OCH2CH3 < CH3OH < CH3CH2OH
Only weak dispersion forces act in CH4 and CH3CH3. The bigger molecule has
more interactions and hence the higher b.p. CH3CH2OCH2CH3 is a bigger
molecule than CH4 and CH3CH3, so has more dispersion forces. It also has
dipole-dipole forces due to the polarised C-O bonds.
CH3OH and CH3CH2OH have hydrogen bonds due to the very electronegative O
atom bonded to the H atom. These H-bonds are much stronger than the
dispersion and dipole-dipole forces in the other compounds and hence these two
compounds have the highest boiling points. CH3CH2OH has more dispersion
forces than CH3OH, so it has the highest boiling point.
• Melting points of the hydrogen halides increase in the order HCl < HBr < HF < HI.
Explain this trend.
There are two competing intermolecular forces at play:
• Dipole-dipole forces increase as the halogen becomes more electronegative (I
< Br < Cl < F).
• Dispersion forces are dependent on the polarisability of the atoms and
increase with the size of the halogen.
Dispersion force dominate in HCl, HBr and HI and determines the order of their
melting points.
The dipole-dipole force in HF is so strong (due to the very small and very
electronegative F atom) that it is given a special name - a hydrogen bond. This
causes HF to have an anomalously high melting point, which just happens to lie
between that of HBr and HI.
2
CHEM1001
2010-J-12
June 2010
• Rationalise the order of the boiling points of the following liquids in terms of their
intermolecular forces.
liquid
F2
HCl
HBr
Cl2
HF
Br2
b.p. (° C)
–188
–85
–67
–34
20
59
The boiling points in F2, Cl2 and Br2 are determined by the size of the dispersion
forces between molecules. The bigger the atoms, the more polarisable their
electron clouds and the greater the dispersion forces. Hence boiling points are in
order Br2 > Cl2 > F2.
Dispersion forces also operate in HF, HCl and HBr, but here the dipole formed
between the halogen atom and the hydrogen also needs to be considered. F is a
very small and very electronegative atom. The H–F bond is therefore highly
polarised and strong H-bonds form in this liquid. These are much stronger than
dispersion forces and so HF has an anomalously high boiling point. Cl and Br are
not as electronegative as F: the dispersion forces in HCl and HBr are more
significant than the dipole-dipole forces as can be evidenced by the order of
boiling points HF > HBr > HCl.
The values given tell us that the total of the dispersion forces in Br2 is greater
than the H-bonds in HF.
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3
CHEM1001
2009-J-4
June 2009
• Sodium chloride is soluble in water, magnesium oxide is not. Using your
understanding of the intermolecular forces involved, explain why this is so.
Both sodium chloride and magnesim oxide are ionic compounds with strong
electrostatic forces between the oppositely charged particles. The energy
required to overcome these attractions is called the lattice enthalpy.
When the ions dissolve in water strong bonds are formed between the ions and
the polar water molecules. The energy released in this process is called the
solvation enthalpy.
If the solvation enthalpy exceeds the lattice enthalpy the compound will be
soluble.
Sodium chloride is soluble because the magnitude of ∆H (i) is less than the
magnitude of ∆H (ii):
(i) NaCl(s) → Na+(g) + Cl–(g)
H2 O
Na+(aq) + Cl–(aq)
(ii) Na+(g) + Cl-(g)
Magnesium oxide is insoluble because the magnitude of ∆H (i) is greater than the
magnitude of ∆H (ii):
(i) MgO(s) → Mg2+(g) + O2–(g)
H2 O
(ii) Mg2+(g) + O2–(g)
Mg2+(aq) + OH–(aq)
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3
CHEM1001
2009-J-11
• The structural formula of acetic acid is shown on the right.
Acetic acid forms dimers (i.e. pairs of molecules) in the gas
phase. Draw the dimer showing the H-bonding that occurs.
June 2009
H
H C C
H
H-bonding
O H
O
H 3C C
C CH3
O
H O
O
O H
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1
CHEM1001
2009-J-12
June 2009
• Which of acetone, (CH3)2CO, and water will have the greater surface tension. Why?
22/01(a)
Marks
2
Water will have the greater surface tension. It has much stronger intermolecular
forces (H-bonds and dispersion) than acetone (dispersion and dipole-dipole
forces).
• Melting points of the hydrogen halides increase in the order HCl < HBr < HF < HI.
Explain this trend.
2
The major intermolecular forces in HCl, HBr and HI are dispersion forces. The
heavier the halogen is, the larger is its electron cloud and the more polarisable it
is. Higher polarisability leads to stronger dispersion forces therefore leading to
melting points which increase in the order HCl < HBr < HI.
F is a very small and very electronegative atom. The H–F bond is therefore
highly polarised and H-bonds form. These are much stronger than dispersion
forces and so HF has an anomalously high melting point. As seen by the
experimental order given, this is enough to raise its melting point above that of
HBr, but not above that of HI.
• Why is the solubility of chloroform (CHCl3) in water 10 times greater than that of
carbon tetrachloride (CCl4) in water?
The C–H bond in CHCl3 is quite polarised due to the electron-withdrawing effect
of the Cl atoms. CHCl3 has a fairly large dipole moment with ô+ H and ô- Cl.
CCl4 has no dipole moment. Although each C-Cl bond is fairly polar, the
tetrahedral shape of the molecule leads to overall cancellation of these bond
dipoles.
The polar water molecules interact better with the polar CHCl3 molecules than
with the non-polar CCl4 molecules so CHCl3 is more soluble.
2
CHEM1001
2008-J-11
June 2008
• Rank the following compounds in order of increasing boiling point? Justify your
answer.
CH3CH2OCH2CH3,
CH3OH,
CH4,
CH3CH3,
CH3CH2OH
CH4 < CH3CH3 < CH3CH2OCH2CH3 < CH3OH < CH3CH2OH
•
CH4 and CH3CH3 have only weak dispersion forces; CH3CH3 has more
electrons so has the dispersion forces are stronger.
•
CH3CH2OCH2CH3 has dispersion forces and, as the presence of the
electronegative oxygen atom leads to a dipole moment, dipole-dipole forces.
•
CH3CH2OH and CH3OH have H attached to electronegative O atoms and so
have relatively strong H-bonds as well as dispersion forces and dipole-dipole
forces. CH3CH2OH has more electrons so has stronger dispersion forces and
hence the higher boiling point.
22/01(a)
3
CHEM1001/CHEM1101 combined
2003-N-12
November 2003
Celcius)
(deg. (C)
pointPoint
Boiling
Boiling
 The figure below shows the boiling points of Group 14 and 17 hydrides as a function
of the period (row) of the periodic table.
0
HF
-50
HI
-100
-150
HBr
HCl
CH4
SiH4
2
3
GeH4
Period
Period
4
SnH4
5
A number of trends are apparent from this figure, including:
- the tetrahydrides have lower boiling points than the monohydrides,
- the boiling point increases with period, with the exception of HF.
Explain these two trends, and the reason that HF is exceptional.
The tetrahydrides are non-polar, so intermolecular attraction is due to
dispersion forces only. As the period increases, the central atom gets bigger
(more electrons) and its polarisability increases - hence the dispersion forces and
boiling points increase.
The monohydrides are polar, so they have dipole - dipole attractions as well as
dispersion forces. Hence they have higher boiling points than corresponding
tetrahydride from same period.
HF is anomalous as the F atom is very small and very electronegative. HF is
therefore able to form strong H-bonds, which are relatively strong
intermolecular attractions. This results in an exceptionally high boiling point for
HF.
Marks
3